(1)ϱíÁгöÁËÈýÖÖÑÀ¸àÖеÄĦ²Á¼Á£¬ÇëÔÚ±íÖÐÌîдÈýÖÖĦ²Á¼ÁËùÊôµÄÎïÖÊÀà±ð£º
| Á½ÃæÕë¶ùͯÑÀ¸à | ÕäÖéÍõ·À³ôÑÀ¸à | ÖлªÍ¸Ã÷ÑÀ¸à |
Ħ²Á¼Á | ÇâÑõ»¯ÂÁ | ̼Ëá¸Æ | ¶þÑõ»¯¹è |
Ħ²Á¼ÁµÄÎïÖÊÀà±ð(Ö¸Ëá¡¢¼î¡¢ÑΡ¢Ñõ»¯Îï) |
|
|
|
(2)¸ù¾ÝÄãµÄÍÆ²â£¬ÑÀ¸àĦ²Á¼ÁµÄÈܽâÐÔÊÇ___________¡£
(3)ÑÀ¸àÖеÄĦ²Á¼Á̼Ëá¸Æ¿ÉÒÔÓÃʯ»ÒʯÀ´ÖƱ¸£¬Ä³Ñ§ÉúÉè¼ÆÁËÒ»ÖÖÖÆ±¸Ì¼Ëá¸ÆµÄʵÑé·½°¸£¬ÆäÁ÷³ÌͼΪ
![]()
![]()
Çëд³öÉÏÊö·½°¸ÖÐÓйط´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
¢Ù____________________________________________________________;
¢Ú____________________________________________________________;
¢Û____________________________________________________________¡£
(4)ÇëÄãÈÔÓÃʯ»ÒʯΪÔÁÏ(ÆäËûÊÔ¼Á×ÔÑ¡)£¬Éè¼ÆÁíÒ»ÖÖÖÆ±¸Ì¼Ëá¸ÆµÄʵÑé·½°¸¡£·ÂÕÕ(3)Ëùʾ£¬½«ÄãµÄʵÑé·½°¸ÓÃÁ÷³Ìͼ±íʾ³öÀ´£º
![]()
ÄãÉè¼ÆµÄ·½°¸ÓŵãÊÇ____________________________________¡£
(5)¼ìÑéÑÀ¸àÖÐÊÇ·ñº¬ÓÐ̼ËáÑεÄʵÑé·½·¨ÊÇ____________________________________¡£
(6)ijѧÉúΪÁ˲ⶨһÖÖÒÔ̼Ëá¸ÆÎªÄ¦²Á¼ÁµÄÑÀ¸àÖÐ̼Ëá¸ÆµÄº¬Á¿£¬ÓÃÉÕ±³ÆÈ¡ÕâÖÖÑÀ¸à¸àÌå100.0 g£¬ÏòÉÕ±ÖÐÖð½¥¼ÓÈëÑÎËáÖÁ²»ÔÙÓÐÆøÌå·Å³ö(³ý̼Ëá¸ÆÍ⣬ÕâÖÖÑÀ¸àµÄÆäËûÎïÖʲ»ÄÜÓëÑÎËá·´Ó¦Éú³ÉÆøÌå)£¬¹²ÊÕ¼¯µ½ÆøÌåµÄÖÊÁ¿Îª22 g¡£ÇëÄã¼ÆËãÕâÖÖÑÀ¸àÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý¡£(¿ÉÄÜÓõ½µÄÏà¶ÔÔ×ÓÖÊÁ¿£ºC¡ª12£»O¡ª16£»Ca¡ª40)
(1)
Ħ²Á¼Á | ÇâÑõ»¯ÂÁ | ̼Ëá¸Æ | ¶þÑõ»¯¹è |
Ħ²Á¼ÁµÄÎïÖÊÀà±ð | ¼î | ÑÎ | Ñõ»¯Îï |
(2)ÄÑÈÜ
(3)¢ÙCaCO3
CaO+CO2¡ü
¢ÚCaO+H2O
Ca(OH)2
¢ÛCa(OH)2+Na2CO3
2NaOH+CaCO3¡ý
(4)![]()
ÓŵãÊÇ£ºÔÚ³£ÎÂÏ·´Ó¦£¬½ÚÊ¡ÄÜÔ´£»±È(3)ÖеIJ½ÖèÉÙ£¬¼õÉÙ²Ù×÷»·½Ú¡£
(5)È¡ÊÊÁ¿ÑÀ¸àÓëÏ¡ÑÎËá·´Ó¦£¬¼ìÑéÊÇ·ñÓжþÑõ»¯Ì¼ÆøÌå²úÉú
(6)ÉèÕâÖÖÑÀ¸à¸àÌåÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊýΪx
CaCO3+2HCl
CaCl2+H2O+CO2¡ü
100 44
100.0x 22g
x=50%
| Äê¼¶ | ¸ßÖÐ¿Î³Ì | Äê¼¶ | ³õÖÐ¿Î³Ì |
| ¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
£¨1£©Ï±íÁгöÁËÈýÖÖÑÀ¸àÖеÄĦ²Á¼Á£¬ÇëÔÚ±íÖÐÌîдÈýÖÖĦ²Á¼ÁËùÊôµÄÎïÖÊÀà±ð¡£
ÑÀ¸à | ij¶ùͯÑÀ¸à | ij·À³ôÑÀ¸à | ij͸Ã÷ÑÀ¸à |
Ħ²Á¼Á | ÇâÑõ»¯ÂÁ | ̼Ëá¸Æ | ¶þÑõ»¯¹è |
Ħ²Á¼ÁµÄÎïÖÊÀà±ð£¨Ö¸Ëá¡¢¼î¡¢Á½ÐÔÇâÑõ»¯Îï¡¢ÑΡ¢Ñõ»¯Î |
|
|
|
£¨2£©¸ù¾ÝÄãµÄÍÆ²â£¬ÑÀ¸àĦ²Á¼ÁµÄÈܽâÐÔÊôÓÚ_______________£¨Ìî¡°Ò×ÈÜ¡±»ò¡°ÄÑÈÜ¡±£©¡£
£¨3£©ÑÀ¸àÖеÄĦ²Á¼Á̼Ëá¸Æ¿ÉÒÔÓÃʯ»ÒʯÀ´ÖƱ¸¡£Ä³Ñ§ÉúÉè¼ÆÁËÒ»ÖÖʵÑéÊÒÖÆ±¸Ì¼Ëá¸ÆµÄʵÑé·½°¸£¬ÆäÁ÷³ÌͼΪ£º
![]()
![]()
Çëд³öÉÏÊö·½°¸ÖÐÓйط´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
¢Ù__________________________________________________________________£»
¢Ú__________________________________________________________________£»
¢Û__________________________________________________________________¡£
£¨4£©ÇëÄãÈÔÓÃʯ»Òʯ×÷ÔÁÏ£¨ÆäËûÊÔ¼Á×ÔÑ¡£©£¬Éè¼ÆÊµÑéÊÒÖÆ±¸Ì¼Ëá¸ÆµÄÁíÒ»ÖÖʵÑé·½°¸£¬ÒÀÕÕ£¨3£©Ëùʾ£¬½«ÄãµÄʵÑé·½°¸ÓÃÁ÷³Ìͼ±íʾ³öÀ´£º
![]()
ÄãÉè¼ÆµÄ·½°¸µÄÓŵãΪ_____________________________________________________¡£
£¨5£©¼ìÑéÑÀ¸àÖÐÊÇ·ñº¬ÓÐ̼ËáÑεÄʵÑé·½·¨ÊÇ_____________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¶ÔÑÀ¸àµÄ̽¾¿ÒªÓõ½Ðí¶à»¯Ñ§ÖªÊ¶¡£
£¨1£©ÏÂÃæÁгöÁËÈýÖÖÑÀ¸àÖеÄĦ²Á¼Á£¬ÇëÔÚ¶ÔӦĦ²Á¼ÁºóÃæºáÏßÉÏÌîдÈýÖÖĦ²Á¼ÁËùÊôµÄÎïÖÊÀà±ð¡£
¢ÙÁ½ÃæÕë¶ùͯÑÀ¸à£ºÇâÑõ»¯ÂÁ £»¢ÚÕäÖéÍõ·À³ôÑÀ¸à£ºÌ¼Ëá¸Æ £»¢ÛÖлªÍ¸Ã÷ÑÀ¸à£º¶þÑõ»¯¹è ¡£
£¨2£©ÑÀ¸àĦ²Á¼ÁµÄÈܽâÐÔÊÇ________________£¨Ìî¡°Ò×ÈÜ¡±»ò¡°ÄÑÈÜ¡±£©
£¨3£©Ä¦²Á¼ÁÇâÑõ»¯ÂÁÊÇ¡°Î¸ÊæÆ½¡±µÄÖ÷Òª³É·Ö£¬¿ÉÒÔÓÃÀ´ÖÎÁÆÎ¸ËᣨÖ÷ÒªÊÇÑÎËᣩ¹ý¶àÖ¢£¬ÇëÄãÓÃÀë×Ó·½³Ìʽ±íʾÆäÔÀí£º________________________________________________________________¡£
£¨4£©Ä¦²Á¼Á¶þÑõ»¯¹èÒ²ÊÇʯӢɰµÄÖ÷Òª³É·Ö£¬ÀûÓÃʯӢɰÓëijÎïÖÊAÔÚµç¯Àï·´Ó¦¿ÉÖÆµÃ´Ö¹è£¨Si£©£¬´Ö¹èÌá´¿ºó¿ÉÖÆµÃ¸ß´¿¶ÈµÄ¹è£¬¿ÉÓÃÓÚÖÆÔìÌ«ÑôÄÜµç³Ø£¬ÄãÈÏΪÎïÖÊAÓ¦¸Ã¾ßÓеÄÐÔÖÊÊÇ____________¡£
A£®Ñõ»¯ÐÔ B£®»¹ÔÐÔ C£®ËáÐÔ D£®¼îÐÔ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¶ÔÑÀ¸àµÄ̽¾¿ÒªÓõ½Ðí¶à»¯Ñ§ÖªÊ¶¡£
£¨1£©Ï±íÁгöÁËÈýÖÖÑÀ¸àĦ²Á¼ÁËùÊôµÄÎïÖÊÀà±ð¡£
| Á½ÃæÕëÑÀ¸à | ÕäÖéÍõ·À³ôÑÀ¸à | ÖлªÍ¸Ã÷ÑÀ¸à |
Ħ²Á¼Á | ÇâÑõ»¯ÂÁ | ̼Ëá¸Æ | ¶þÑõ»¯¹è |
Ħ²Á¼ÁµÄÎïÖÊÀà±ð £¨Ö¸Ëá¡¢¼î¡¢ÑΡ¢Ñõ»¯Î |
|
|
|
£¨2£©¸ù¾ÝÄãµÄÍÆ²â£¬ÑÀ¸àĦ²Á¼ÁµÄÈܽâÐÔÊÇ__________£¨Ìî¡°Ò×ÈÜ¡±¡°ÄÑÈÜ¡±£©¡£
(3) ijͬѧÏëÖ»ÓÃÑÎËáÀ´¼ìÑéÄ³Ò»Æ·ÅÆµÄÑÀ¸àµÄÄ¥²Á¼ÁÊÇÉÏÊöÈýÖÖÖеÄÄÄÒ»ÖÖ£¬ÇëÎÊÓ¦¸ÃÈçºÎ²Ù×÷£¬²¢Ð´³öÓйط´Ó¦µÄÀë×Ó·½³Ìʽ¡£____________________________________________________________________________________________
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
(9·Ö)¶ÔÑÀ¸àµÄ̽¾¿ÒªÓõ½Ðí¶à»¯Ñ§ÖªÊ¶¡£
(1)ϱíÁоÙÁËÈýÖÖÑÀ¸àÖеÄĦ²Á¼Á£¬ÇëÔÚ±íÖÐÌîдĦ²Á¼ÁËùÊôµÄÎïÖÊÀà±ð¡£
ÑÀ¸à | Á½ÃæÕë¶ùͯÑÀ¸à | ÕäÖéÍõ·À³ôÑÀ¸à | ÖлªÍ¸Ã÷ÑÀ¸à |
Ħ²Á¼Á | ÇâÑõ»¯ÂÁ | ̼ÀÒ¸Æ | ¶þÑõ»¯¹è |
ÎïÖÊÀà±ð | Á½ÐÔÇâÑõ»¯Îï | ¢Ù | ¢Ú |
(2)¸ù¾ÝÄãµÄÍÆ²â£¬ÑÀ¸àĦ²Á¼ÁµÄË®ÈÜÐÔÊôÓÚ (Ìî¡°Ò×ÈÜ¡±»ò¡±ÄÑÈÜ¡±)¡£
(3)Ħ²Á¼ÁÇâÑõ»¯ÂÁÊÇ¡°Î¸ÊæÆ½¡±µÄÖ÷Òª³É·Ö£¬¿ÉÒÔÓÃÀ´ÖÎÁÆÎ¸Ëá(Ö÷ÒªÊÇÑÎËá)¹ý¶àÖ¢£¬ÇëÄãÓÃÀë×Ó·½³Ìʽ±íʾÆäÔÀí ¡£
(4)Ħ²Á¼Á̼Ëá¸Æ¿ÉÒÔÓÃʯ»ÒʯÃ×ÖÆ±¸£¬Ä³Ñ§ÉúÉè¼ÆÁËÒ»¸öÖÆ±¸Ì¼Ëá¸ÆµÄʵÑé·½°¸£¬ÆäÁ÷³ÌͼΪ£º
![]()
ÇëÄãÈÔÓÃʯ»Òʯ×÷ÔÁÏ(ÆäËûÊÔ¼Á×ÔÑ¡)£¬Éè¼ÆÁíÒ»ÖÖÖÆ±¸Ì¼Ëá¸ÆµÄʵÑé·½°¸¡£°´ÕÕÉÏÀýËùʾ£¬½«ÄãµÄÁ÷³Ìͼ±íʾ³öÀ´£º
![]()
(5)Ħ²Á¼Á¶þÑõ»¯¹èÒ²ÊÇʯӢɰµÄÖ÷Òª³É·Ö£¬ÀûÖÝʯӢɰÓ뽹̿ÔÚµç¯Àï·´Ó¦¿ÉÖÆµÃ´Ö¹è£¬´Ö¹èÌá´¿ºó¿ÉÖÆµÃ¸ß´¿¶ÈµÄ¹è(Si)¡£Ð´³öʯӢɰÓ뽹̿ÔÚµç¯ÀïÖÆµÃ´Ö¹èµÄ»¯Ñ§·½³Ìʽ ¡£
²é¿´´ð°¸ºÍ½âÎö>>
¹ú¼ÊѧУÓÅÑ¡ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com